Physics 8

Spring 2012

NAME:

TA:

Physics 8 Spring 2012 Midterm 2 Solutions For the midterm, you may use one sheet of notes with whatever you want to put on it, front and back. Please sit every other seat, and please don’t cheat! If something isn’t clear, please ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR FINAL ANSWERS! You have the full length of the class. If you attach any additional scratch work, then make sure that your name is on every sheet of your work. Good luck! 1. While our Sun doesn’t have enough mass to ever become a black hole, suppose that it could. (a) What would be the Schwarzschild radius of the Sun? (b) Using Kepler’s law, determine the orbital period in years of the Earth around this new black hole. Hint: for this problem you might find the following information useful: GN = 6.672 × 10−11 N m2 /kg2 , MSun = 2 × 1030 kg, MEarth = 5.97 × 1024 kg, c = 2.998 × 108 m/s, and the orbital radius of the Earth is 1.5 × 1011 meters. ———————————————————————————————————— Solution (a) The Schwarzschild radius of the Sun is given by RS =

2GN MSun . c2

Plugging in the values gives RS =

2GN MSun 2 × 6.672 × 10−11 × 2 × 1030 = = 2970 meters, c2 (2.998 × 108 )2

or about 3 kilometers. (b) Kepler’s law relates the period of the orbit to its radius, r, by 4π 2 r3 . GN MSun Since the mass of the Sun didn’t change upon becoming a black hole, the orbital period is exactly the same as it always was, namely one year. We can check this just by plugging in the numbers s s 2 4π 4π 2 × (1.5 × 1011 )3 T = r3 = = 3.16 × 107 seconds, GN MSun 6.672 × 10−11 × 2 × 1030 T2 =

which is a year! 1

2. Bored, a boy shoots his pellet gun at a piece of cheese that sits on a massive block of ice. On one particular shot, his 1.2 g pellet gets stuck in the cheese, causing it to slide 25 cm before coming to a stop. If the muzzle velocity of the gun is known to be 65 m/s and the cheese has a mass of 120 g, what is the coefficient of friction between the cheese and the ice? ———————————————————————————————————— Solution Let’s start by looking at the momentum. Since the cheese isn’t moving, initially, the initial momentum is just that of the pellet, pi = ppellet . If the mass of the pellet is m, and it has an initial velocity v, then pi = mv. After the collision, the pellet is stuck in the cheese. If the cheese has a mass M , and if the system has a final velocity V , then the final momentum is pf = (m + M ) V . Solving for the final velocity gives   m v. V = m+M So, the combined system has an initial kinetic energy of KE = 12 (m + M ) V 2 = m2 v 2 . After the cheese/pellet system has come to rest, its kinetic energy is zero. 2(m+M ) The energy lost has to go into the work done by friction, which is W = µk FN d = µk (m + M ) gd, where d is the distance the system slides. So, setting this equal to the kinetic energy and solving for the coefficient we find µk =

m2 v 2 . 2 (m + M )2 gd

Plugging in numbers gives µk =

m2 v 2 0.00122 × 652 = = 0.085 2 (m + M )2 gd 2 (0.1212)2 × 9.8 × 0.25

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3. According to the Standard Model of Particle Physics, electrons are pointlike particles having no spatial extent. (This assumption has been confirmed experimentally, and the radius of the electron has been shown to be less than 10−18 meters.) The intrinsic spin of an electron could in principle be due to its rotation. Let us check to see if this conclusion is feasible. (a) Assuming that the electron is a uniform sphere whose radius is 1.00 × 10−18 m, what angular speed would be necessary to produce the observed intrinsic angular momentum of ~/2? (Hint: Recall that the moment of inertia of a solid sphere is I = 52 M R2 , ~ = 1.05 × 10−34 J s, and that the mass of an electron is 9.11 × 10−31 kg.) (b) Using this value of the angular speed, show that the speed of a point on the “equator” of a “spinning” electron would be moving faster than the speed of light (which Einstein tells us is impossible). What is your conclusion about the spin angular momentum being analogous to a spinning sphere with spatial extent? ———————————————————————————————————— Solution (a) The angular momentum can be expressed in terms of the moment of inertia, I, and the angular velocity, ω, as L = Iω. Thus, the angular speed is ω = L/I. Now, for a uniform sphere, I = 25 M R2 , and for L = ~/2, we find ω=

5~ L = . I 4M R2

Taking M = me = 9.11 × 10−31 kg, we have ω=

5 × 1.05 × 10−34 5~ = = 1.44 × 1032 rad/sec. 4M R2 4 × 9.11 × 10−31 × (10−18 )2

(b) The speed is v = Rω, so using our results from part (a) gives v = Rω = 10−18 × 1.44 × 1032 = 1.44 × 1014 m/s! This is much faster than light. So, if we require that these “points” on the “equator” can’t spin faster than light, we have to abandon the model of the electron as a tiny little spinning sphere and think of spin in a different way.

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4. A basketball rolls without slipping down an incline of angle θ. The coefficient of static friction is µs . Model the ball as a thin spherical shell. Find (a) the acceleration of the center of mass of the ball, (b) the frictional force acting on the ball from the torque, and

(c) the maximum angle of the incline for which the ball will roll without slipping (this Picture the (b) Problem The Fthree happens when the friction from part just equals f = µforces s FN ). acting on the basketball are

the ball, the normal force, and the force of friction. Because the assumed to be acting at the center of mass, and the normal force a ———————————————————————————————————— center of mass, the only force which exerts a torque about the ce the frictional force. Let the mass of the basketball be m and apply Solution law to find a system of simultaneous equations that we can solve fo (a) We can begin by writing down Newton’s laws for called for in the problem statement.

Hint: The moment of inertia of a spherical shell is I = 23 mR2 .

the ball, including the torques: P P Fx = −mg sin θ + Ff = −ma 0 PFy = −mg cos θ + Fn = τi = Ff R = Iα.

y

Furthermore, since the ball is rolling without slipping, we have that a = αR. So, we have enough to solve this system of equations. Plugging in Ff = Iα/R = Ia/R2 to the first equation and solving for a gives

m x

r Fn r 0

r mg θ

mg sin θ . m + I/R2 (a) Apply Newton’s 2nd law in2both2 The moment of inertia of the spherical shell is just I = 3 mR , and so a=

translational and rotational form to mg sin θ 3 thea ball: = = g sin θ. 2 m + 2m/3R

(b) Since Ff =

I a R2

=

2 mR2 3 R2

5

a = 23 ma, we have 2

mg sin θ. is rolling BecauseFthe f =basketball 5 without slipping we know that:

r f

∑F ∑F

x

= mg sin θ − f s = m

y

= Fn − mg cosθ =

and

∑τ

α=

= f s r = I 0α

0

a r

(c) Now, we know that the frictional force is Ff = µs FN = µs mg cos θ. Setting this equal to our result from part (b) we find

a

Substitute in equation (3) to

fs r = I 0 2 5 mg sin θ = µs mg cos θ ⇒ tan θ = µs . r obtain: 5 2 So, the maximum angle of the incline that the ball will roll without slipping is 2 From Table 9-1 we I 0 = 23 mr  have:  5 θ = tan−1 µs . 2

Substitute for I0 and α in equation (4) and solve for fs:

The rolling of the ball increases the angle from the sliding case.

Substitute4for fs in equation (1) and solve for a:

fs r = a=

(

2 3

3 5

g sin θ

mr 2

) ar ⇒ f

s

=

2 3

Extra Credit Question!! The following is worth 10 extra credit points! Some people think that the shuttle astronauts are “weightless” because they are “beyond the pull of Earth’s gravity.” In fact, this is completely untrue. (a) What is the magnitude of the gravitational field in the vicinity of a shuttle orbit? A shuttle orbit is about 400 km above the ground. (b) Given the answer in Part (a), explain why shuttle astronauts suffer from adverse biological effects such as muscle atrophy even though the are not actually “weightless.” ———————————————————————————————————— Solution (a) The shuttle orbits at a distance of r = RE + h from the center of the Earth, where RE is the radius of the Earth, and h is the height above the surface. If the height is h = 400 km, then r = 6400 + 400 = 6800 km, or 6.8 × 106 meters. At this point, the gravitational field has a magnitude 6.67 × 10−11 × 5.98 × 1024 GME = = 8.6 m/s2 , g= 2 6 2 r (6.8 × 10 ) which is still almost 90% of the acceleration at the surface of the Earth. (b) Remember that the weight that we feel is due to the normal force. The astronauts in orbit are in constant free-fall, and don’t feel their weight. So, without the compensating normal force to fight against the muscles begin to weaken, not needing to do as much anymore.

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